I understand that if you don't know the trick, it's not easy create a solution with a complexity of \$O(N)\$. However, I would like to ask you how to solve this tricky question:
You are given two non-empty zero-indexed arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river. The fish are numbered from 0 to N − 1, fish number P is represented by
A[P]
andB[P]
, and if P < Q then fish P is initially upstream of fish Q. Initially, each fish has a unique position. Array A contains the sizes of the fish. All its elements are > unique. Array B contains the directions of the fish. It contains only 0s and/or 1s, where:0 represents a fish flowing upstream 1 represents a fish flowing downstream
If two fish move in opposite directions and there are no other (living) fish between them, they will eventually meet each other. Then only one fish can stay alive − the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P < Q,
B[P]
= 1 andB[Q]
= 0, and there are no living fish between them. After they meet:If A[P] > A[Q] then P eats Q, and P will still be flowing downstream, If A[Q] > A[P] then Q eats P, and Q will still be flowing upstream.
We assume that all the fish are flowing at the same speed. That is, fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.
For example, consider arrays A and B such that:
A[0] = 4 B[0] = 0 A[1] = 3 B[1] = 1 A[2] = 2 B[2] = 0 A[3] = 1 B[3] = 0 A[4] = 5 B[4] = 0
Initially all the fish are alive and all except fish number 1 are moving upstream. Fish > number 1 meets fish number 2 and eats it, then it meets fish number 3 and eats it too. Finally, it meets fish number 4 and is eaten by it. The remaining two fish, numbers 0 and 4, never meet and therefore stay alive.
Write a function:
class Solution { public int solution(int[] A, int[] B); }
that, given two non-empty zero-indexed arrays A and B consisting of N integers, returns the number of fish that will stay alive.
For example, given the arrays shown above, the function should return 2, as explained above.
Assume that:
N is an integer within the range [1..100,000], where each element of array A is an integer within the range [0..1,000,000,000], where each element of array B is an integer that can have one of the following values: 0, 1, where the elements of A are all distinct.
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
My solution:
import java.util.Stack;
class Solution {
public int solution(int[] A, int[] B) {
int n = A.length;
if (n<1 || n>100000) return -1;
int alive = n;
Stack<Integer> down = new Stack<Integer>();
// array to check if a fish has died
boolean[] lives = new boolean[n];
for (int i=0;i<n;i++) lives[i]=true;
// the first fish with 1 is going to the right and will try
// to eat other fishes in opposite direction
// but if the fish is bigger it will comeback and it will eat other fishes in opposite direction
for (int i=0,k=i+1; i<n-1;i++) {
if (B[i]==0) continue;
for (;k<n;k++){
// I save in a stack all the fishes in the same direction so in this way
//I don't have to check again back if there is some fish
//that is bigger than the fish that has been capable to eat the "i" fish
if (B[i]==B[k] && B[i]==1) down.push(k);
if (A[i]>A[k] && B[i]!=B[k] && B[i]==1 && lives[k]){
alive--;
lives[k]=false;
} else if (A[i]<A[k]&&B[i]!=B[k]&&B[i]==1 && lives[i]) {
alive--;
lives[i]=false;
//k fish eat i fish and I try to check if exists some i+1 to i-1
//fish that is opposite to k fish and could eat "k" fish
while(!down.empty() && i!=down.peek()){
i=down.pop();
if (A[i]<A[k] && lives[i]) {
alive--;
lives[i]=false;
} else if (A[i]>A[k] && lives[k]){
alive--;
lives[k]=false;
break;
}
}
}
}
}
return alive;
}
}
This solution is not completely right! I found the right solution but in \$O(N^2)\$ time. I was trying to reach \$O(N)\$. If you want to test, check this URL.